Question

The value of c in Rolle's theorem for the function $f\left(x\right)=\frac{x\left(x+1\right)}{e^x}$ defined on [−1, 0] is
(a) 0.5

(b) $\frac{1+\sqrt{5}}{2}$

(c) $\frac{1-\sqrt{5}}{2}$

(d) −0.5

Answer 1

(c) $\frac{1-\sqrt{5}}{2}$1−5√2

Given:
$f\left(x\right)=\frac{x\left(x+1\right)}{e^x}$

Differentiating the given function with respect to x, we get

$$\begin{gathered} f\text{'}\left(x\right)=\frac{e^x\left(2x+1\right)-x\left(x+1\right)e^x}{{\left(e^x\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{\left(2x+1\right)-x\left(x+1\right)}{e^x} \\ \Rightarrow f\text{'}\left(x\right)=\frac{2x+1-x^2-x}{e^x} \\ \Rightarrow f\text{'}\left(x\right)=\frac{-x^2+x+1}{e^x} \\ \Rightarrow f\text{'}\left(c\right)=\frac{-c^2+c+1}{e^c} \\ \therefore f\text{'}\left(c\right)=0 \\ \Rightarrow \frac{-c^2+c+1}{e^c}=0 \\ \Rightarrow c^2-c-1=0 \\ \Rightarrow c=\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2} \\ \therefore c=\frac{1-\sqrt{5}}{2}\in \left(-1,0\right) \end{gathered}$$


Hence, the required value of c is $\frac{1-\sqrt{5}}{2}$.